The Two Envelope Paradox in Probability and Expected Return
Date: 09/20/2005 at 16:11:24 From: Mike Subject: Probability, Expected Return Consider this problem. I truthfully tell you I have money in each of my hands concealed behind my back, and one hand contains half the money of the other. I let you choose one hand, and then give you the money in that hand. I then give you the option of either keeping the money, or buying the contents of the other hand with the money you just received. It's clear that neither choice offers an advantage. Half the time your first choice yields the larger amount, and half the time it yields the smaller. Switching doesn't change the expected outcome over a large number of trials. But letís pretend that we don't know which choice offers an advantage, and decide to use the expected return of each outcome to judge which is better. Letís say the hand picked yields $1. The expected return of keeping the first hand is $1. If you choose to buy the contents of the other hand, the expected return is Ĺ x $2 + Ĺ x $0.50, because the other hand may contain either $2 or $0.50, and either choice is equally likely. The result of that expression is $1.25. Since $1.25 is greater than $1, the implication is clear that over a large number of trials the preferred strategy is to buy the other hand. But we previously concluded that neither strategy is superior. Why does the expected return calculation yield bogus results? The expected return calculation is apparently performed correctly, but the result appears to be incorrect.
Date: 09/21/2005 at 11:06:37 From: Doctor Julien Subject: Re: Probability, Expected Return Hello Mike, This is quite possibly my favorite problem! It's an often misunderstood problem that is much deeper than it appears to be. A thorough analysis can be found here: The Two-Envelope Paradox: A Complete Analysis? http://www.consc.net/papers/envelope.html It might be a little bit of a rough read--here is another, easier to read one: Two Envelopes Paradox http://www.math.hmc.edu/funfacts/ffiles/20001.6-8.shtml Please feel free to write back if you still have questions about this problem, or need help in understanding something. - Doctor Julien, The Math Forum http://mathforum.org/dr.math/
Date: 09/22/2005 at 11:07:05 From: Mike Subject: Probability, Expected Return Doctor Julien, Thanks for your reply. I read both papers you supplied. The "thorough analysis" is beyond me, and I'm not sure I fully understand the "easier to read one". First, it uses the term "condition" in a way I don't understand. It then says, "You need to have some idea of what the prior distribution of money in the envelope is before you can do the calculation" (without saying why), and goes on to show that everything works fine when we know what is in each envelope. Is this saying that the probability in the expected return calculation (i.e. 0.5 for each outcome) is wrong because we don't really know the probability of each value? If so, I'd think that we could alter the problem slightly so we did know the probability and still have the same paradox. Say that the process used by the offerer was to select the first value randomly, and then select a second value that is either 2 times that value, or Ĺ times that value, with equal probability. That would ensure that we DID know that probability was Ĺ. If we ensure that the probabilities are correct (and we know the return is either 2x or 1/2x from the problem statement), how can the distribution make a difference? Does my statement make sense, or am I totally missing the argument? Thanks, Mike
Date: 9/23/2005 at 13:56:34 From: Doctor Julien Subject: Re: Probability, Expected Return Hello Mike, There are several layers to this argument. In the example you provided, you said "select the first value randomly". The problem is, how do you do that? If we say that you pick any integer with equal probability, p, then we have an infinite number of possibilities each having the same probability p. This is impossible since the sum of all probabilities must sum to 1. Thus there is no way to randomly and uniformly pick a number. That is why the "easier to read" paper talks about having to know the distribution. However, there _is_ a way to select random values that will work. For each envelope, a coin is flipped until it comes up heads, and if it came up heads on the nth trial, 2^n is put into the envelope. If we do this, then the sum of probabilities is: 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + .... = 1 This situation is discussed here: The St. Petersburg Two-Envelope Paradox http://www.consc.net/papers/stpete.html The argument that the author makes is that the expected value of each envelope is infinite. Thus, if you open an envelope with a finite amount of money, you should always want to switch. So even if you saw a billion dollars in the envelope, you should want to switch because there could be so much more in the other. The paradox is that the situation is symmetrical! So by symmetry you shouldn't want to switch. As the author puts it: When a distribution over finite amounts has an infinite expected value, any specific result will be disappointing. It will then always be in one's interest to do things over, if given the opportunity. But it does not follow from this that before knowing the result, it is in one's interest to do things over. I recommend reading the paper. It's by the same author as the earlier "hard" one, but this one doesn't have any math. If this doesn't answer your question, or if you still have difficulty, please write back! - Doctor Julien, The Math Forum http://mathforum.org/dr.math/
Date: 9/26/2005 at 12:03:49 From: Mike Subject: Probability, Expected Return Doctor Julien, Thanks for your response. I read the paper you provided. I had to skip over some parts, but as I read the remainder of the paper I understood some fragments but not the whole. I think the upshot of all this is that there's too much math I don't know for me to be successful here. I could probably (I think) do it, but it would take more effort than I have time to apply to this. But maybe I can understand a simpler question: in what situations does the straightforward use of expected returns fail to provide a correct result? By reading your comments and the paper you provided it seems to me that there are two cases that fail: 1) The probability of each outcome can't be computed. The example is the original problem, where as you point out "there is no way to randomly and uniformly pick a number" because there are an infinite number of numbers to chose from. 2) The expected value of the value of an outcome is infinite. An example is where "a coin was flipped until it came up heads, and if it came up heads on the nth trial, 2^n is put into the envelope." Generalizing, failure occurs when the expected value of any outcome is infinite. Is my understanding correct? Thanks for taking the time to explain this to me. Regards, Mike
Date: 9/27/2005 at 16:43:32 From: Doctor Julien Subject: Re: Probability, Expected Return Hello Mike, Yes, you almost have it with your understanding. The author says: The upshot is a disjunctive diagnosis of the two-envelope paradox. The expected value of the amount in the envelopes is either finite or infinite. If it is finite, then (1) and (2) are false: the paradoxical reasoning results from illicitly ignoring probability distributions. If it is infinite, then the step from (2) to (3) is invalid: the paradoxical reasoning relies on dominance principles or probabilistic principles that are false in the infinite case.[*] What he is saying is that there are two cases: 1) The expected value in the envelope is finite. If this is the case, then we are ignoring probability distributions. Our decision to switch or not to switch will change based on the amount we open in our own envelope and what we know about the probability distribution and expectation. 2) The expected value in the envelope in infinite. If this is the case, then no matter what amount we open, we will want to switch. However, it does not follow that we want to switch before opening the envelope! This is what happens when you have infinite expectation! Any specific result will be disappointing. Does this help clear things up? - Doctor Julien, The Math Forum http://mathforum.org/dr.math/
Date: 10/02/2005 at 13:17:46 From: Mike Subject: Thank you (Probability, Expected Return) Dr. Julien, I want to thank you for making the effort to discuss this in a manner I could understand. After reading the papers you suggested I thought it was beyond my grasp given the level of my math training, but now I think I understand it as well as I can short of going back to college! Thanks again for your help. Regards, Mike
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